A) \[\frac{1}{\sqrt{2}}\]
B) \[\frac{1}{2\sqrt{2}}\]
C) \[\frac{-\,1}{\sqrt{2}}\]
D) \[-\frac{1}{2\sqrt{2}}\]
Correct Answer: A
Solution :
| We have, |
| \[\left| f(x) \right|+\sqrt{1+\cos 2\pi x}={{\tan }^{2}}\left( \frac{\pi x}{9} \right)f(x)\] |
| At \[x=3\] |
| \[\left| f(3) \right|+\sqrt{1+\cos 6\pi }={{\tan }^{2}}\left( \frac{\pi }{3} \right)f(3)\] |
| \[\left| f(3) \right|+\sqrt{2}=3f(3)\] |
| If \[f(3)\ge 0\]\[\Rightarrow \]\[f(3)=\frac{1}{\sqrt{2}}\] |
| If \[f(3)<0\]\[\Rightarrow \]\[f(3)=\frac{\sqrt{2}}{4}\]not possible. |
| Hence, \[f(3)=\frac{1}{\sqrt{2}}\] |
You need to login to perform this action.
You will be redirected in
3 sec
